Math question - probability

[NuclearSilo]

Let's say you have 6 numbers: 1 2 3 4 5 6

1st case:
-1st attempt: you can choose 2 numbers among these 6
-2nd attempt: choose 1 number among the 2 numbers you just pick

2nd case:
only one attempt is to choose 1 number among 6

Question: what's the probability to pick the number 1 and does both case has the same probability to pick a number?

[rumpleKillskin]

Why do I have the feeling that this is coming out of your homework..

[Barotix]

Wouldn't it stay 1/6th assuming they don't repeat? oO Or is this some "really hard math" that I don't know of yet.

[NO_SILK_4_ME]

Wouldn't it stay 1/6th assuming they don't repeat? oO Or is this some "really hard math" that I don't know of yet.

That.

OT: You should come back to ecSRO

[Bop]

if this is all in randomness, then 1/6*1/2=1/12 for case 1 and 1/6 for case 2 <.<

not a mathematician...

[Aaron777]

if this is all in randomness, then 1/6*1/2=1/12 for case 1 and 1/6 for case 2 <.<

not a mathematician...

No, it would be 1/3 (2/6) because you picked 2 of the 6 in the first attempt.. so then it would come out to equal 1/3(1/2) =1/6 =situation 2

[Stephanus]

Well this reminds me of a book i received from a random yehova's witnesses encounter(no, they gave it to me, it wasnt a loot :p )
That "book" had similar examples which tried to convince me, god is behind probability.
Looking at this... im starting to believe.
So i refuse to answer based on my religions laws. Only god knows.


ps: if you think about this.
1 is 1 becoz ppl told you its 1.
Also 2 is 2 becoz you were told its 2.
If you think more, math is based on belief and its a religion wich hides himself behind false logic!!!!1111!!!!444

[Goobronicus]

Well...you never say you are drawing blindly or at random, so if it's a trick question then ...

Otherwise yes, they have the same probability. Sequential = multiply (1/3 x 1/2; instead of 1/3 +1/2)

[woutR]

Let's say you have 6 numbers: 1 2 3 4 5 6

1st case:
-1st attempt: you can choose 2 numbers among these 6
-2nd attempt: choose 1 number among the 2 numbers you just pick

2nd case:
only one attempt is to choose 1 number among 6

Question: what's the probability to pick the number 1 and does both case has the same probability to pick a number?

first case, first attempt, 2/6 chance to get your desired number = 1/3.
2nd attempt 0.5 chance to get the number you wanted, so 1/3 x .5= 0,17

Choosing one number is pretty easy mode, it's 1/6. With each more choice you get the chance goes up 1/6.

But anyways, I think this is easy math leading up to real probability maths with things like there are 3 red and 7 blue balls, you can pick 5, how big is the chance you get 3 red and 2 blue?
But that's also not very hard.

[Brago]

Ikno this is randomly off topic but
my teacher said that there is only one actual number and that is the number 1... the other numbers are basically symbols.

2 = 2 1's
etc..

[only_me]

Let's say you have 6 numbers: 1 2 3 4 5 6

1st case:
-1st attempt: you can choose 2 numbers among these 6
-2nd attempt: choose 1 number among the 2 numbers you just pick

the probability to pick the number 1 is 1/6 ~ 16%

2nd case:
only one attempt is to choose 1 number among 6

is the same probability to pick it : 1/6 ~ 16%

Question: what's the probability to pick the number 1 and does both case has the same probability to pick a number?

yes , the probability is the same

[woutR]

I don't really agree the chance is 1/6.
I can understand how you can think that because you half the 2/6 with picking only 1 out of two, leading you back to 1/6. But I've been learned you should multiply the probabilities in this case.

Just like with tests, like the first time to succeed your test the probability is .8 and then when you take it again it's .5. The chance will not be .4 but .8 x .5

[only_me]

I don't really agree the chance is 1/6.
I can understand how you can think that because you half the 2/6 with picking only 1 out of two, leading you back to 1/6. But I've been learned you should multiply the probabilities in this case.

Just like with tests, like the first time to succeed your test the probability is .8 and then when you take it again it's .5. The chance will not be .4 but .8 x .5

-1st attempt: you can choose 2 numbers among these 6
-2nd attempt: choose 1 number among the 2 numbers you just pick

this is just a trick ... it doesn't affect the propability .
in the end u will have just 1 number

[NuclearSilo]

Oh, ok. But somehow, i have the feeling that the 1st case has more chance...

Example:

You have 1 million of different number. Randomly pick one. What's the chance of getting the number 123?

vs

You have 1 million of different number. For each attempt you choose half of the number you just picked. What's the chance of getting the number 123?

You'll notice that with the 2nd method, sooner or later you'll get the number 123. But with the 1st method, maybe you got to choose for unlimited of times but still not got the wanted number.
Yes, yes, I know. You are telling me in math it's the same, but sadly in real life, it's not.

[woutR]

I don't really agree the chance is 1/6.
I can understand how you can think that because you half the 2/6 with picking only 1 out of two, leading you back to 1/6. But I've been learned you should multiply the probabilities in this case.

Just like with tests, like the first time to succeed your test the probability is .8 and then when you take it again it's .5. The chance will not be .4 but .8 x .5

-1st attempt: you can choose 2 numbers among these 6
-2nd attempt: choose 1 number among the 2 numbers you just pick

this is just a trick ... it doesn't affect the propability .
in the end u will have just 1 number

Well, you're actually right. My method assumes you picked the right number in your first 2/6 try, which they don't say is the case.
So the chance is indeed equal.

[Bop]

Oh, ok. But somehow, i have the feeling that the 1st case has more chance...

Example:

You have 1 million of different number. Randomly pick one. What's the chance of getting the number 123?

vs

You have 1 million of different number. For each attempt you choose half of the number you just picked. What's the chance of getting the number 123?

You'll notice that with the 2nd method, sooner or later you'll get the number 123. But with the 1st method, maybe you got to choose for unlimited of times but still not got the wanted number.
Yes, yes, I know. You are telling me in math it's the same, but sadly in real life, it's not.

real value will still be close to expected value. eg. flipping a coin 100 times, u may get 48 heads, 52 tails, u can use the chi-square or w/e to find accuracy

[Azilius]

1st case:
-1st attempt: you can choose 2 numbers among these 6
-2nd attempt: choose 1 number among the 2 numbers you just pick

So I replace 1 with a, 2 with b, 3 with c, 4 with d, 5 with e, and 6 with f

I pick b and c, neither are #1
Why the hell would I pick from those two again when on the first attempt I've been told, neither are #1??


I'm not sure if the wording is off or I just don't understand, but on the first 'attempt' of the first case, if neither was the number you needed, why the hell would you pick from those two again?


Also..


Your second example does not work. You'd have to chose a number that eventually halves to 123 to be able to get the number 123. If I picked 1,000,000, then it's half, I'd get 500,000. Then i'd get 250,000. Then 125,000, then 62,500, then 31,250, then 15,625, etc etc etc. If you had worded it differently and made it so you'd get the number 123 eventually, the 'chance' would be 100% because you'd always get 123.

[NuclearSilo]

Why the hell would I pick from those two again when on the first attempt I've been told, neither are #1?

Well i don't think there is a need to answer this question because the goal here is to answer to the probablity question, and not the WHY.

In the 2nd example, you choose a list of numbers, not just one number.
You have 10k numbers, you divide it in 2: 0->4999, 5k->10k
You see if the wanted is lower than the half, you take the lower part, if higher, take higher part. The choice of the part is random, if it's the right part you continue the choosing process, if it's not, do it from the beginning. Do until you have only one number and it's the wanted number.

[Grimjaw]

Well this reminds me of a book i received from a random yehova's witnesses encounter(no, they gave it to me, it wasnt a loot :p )
That "book" had similar examples which tried to convince me, god is behind probability.
Looking at this... im starting to believe.
So i refuse to answer based on my religions laws. Only god knows.


ps: if you think about this.
1 is 1 becoz ppl told you its 1.
Also 2 is 2 becoz you were told its 2.
If you think more, math is based on belief and its a religion wich hides himself behind false logic!!!!1111!!!!444

But what if you're a atheist? Yet you are still able to solve the math you are being presented with. Wouldn't the atheist factor implemented then overthrow you're theory? And send those Jehovah's to the moon where they belong? Since you say you started to believe God determines probability, whereas I say God has nothing to do with solving mathematical questions or determining probability or grant some insight into equations. If I could choose my faith freely, which I can, I wouldn't count on God's inspiration or mathematical touches when it came to solving the puzzles.

[Azilius]

Why the hell would I pick from those two again when on the first attempt I've been told, neither are #1?

Well i don't think there is a need to answer this question because the goal here is to answer to the probablity question, and not the WHY.

In the 2nd example, you choose a list of numbers, not just one number.
You have 10k numbers, you divide it in 2: 0->4999, 5k->10k
You see if the wanted is lower than the half, you take the lower part, if higher, take higher part. The choice of the part is random, if it's the right part you continue the choosing process, if it's not, do it from the beginning. Do until you have only one number and it's the wanted number.

It wasn't a 'why' question, it was for clarification. You've explained the second thing for me nicely but your broken English hasn't explained the first. My point was, going by how I read it, the question didn't make sense.

[Grimjaw]

Lol. You can't take 70% of the things here serious, if you did, you would quickly find out that there are almost 0 topics left which are interesting enough to post in.

[NuclearSilo]

It wasn't a 'why' question, it was for clarification. You've explained the second thing for me nicely but your broken English hasn't explained the first. My point was, going by how I read it, the question didn't make sense.

The first and the second are almost the same. If other members understood, why can't you?

[Plutonium]

apparently you are not very intelligent at math NuclearSilo... your answers dont make sense lol.

who on SRF is a math whiz that will be able to answer this?

[NuclearSilo]

Which answers? I didn't answer anything. I was asking for the answer lol

[Plutonium]

:?
Which answers? I didn't answer anything. I was asking for the answer lol

like many have stated, it's obvious the answer is 1/6 for both. it's a 2nd grade probability question. why are you disagreeing with people if u are asking a question that u dont know?

anways...

[NuclearSilo]

No, you got it wrong. I agree that the answer is 1/6. I just don't agree with other things.

[Plutonium]

No, you got it wrong. I agree that the answer is 1/6. I just don't agree with other things.

other things being what?

[pipigrande]

Let's say you have 6 numbers: 1 2 3 4 5 6

1st case:
-1st attempt: you can choose 2 numbers among these 6
-2nd attempt: choose 1 number among the 2 numbers you just pick

2nd case:
only one attempt is to choose 1 number among 6

Question: what's the probability to pick the number 1 and does both case has the same probability to pick a number?

I will answer this question. This comes from a 2nd year Biophysics student, but I believe a grade 12 class in probability&statistics would be enough.

Here is how I would solve the questions:

1st case:
(2/6 x 1/2) = 0.166. There is ~17 that you would pick the number 1.

2nd case:
1/6 = 0.166. There is ~17 that you would pick the number 1.

Both cases give you the same probability of picking the number 1 or any number for that matter.

PS: For people that can't understand why both give you same answer, I will try to explain it here:

In the first case, when choosing 2/6, you are basically choosing 1/3 and therefore have 33% of choosing the number 1. Suppose you have chosen the right number '1', then the probablity now of picking the right number is 50%. However, the answer is not 50%. You gotta multiply the base rate (33%) by the second probability (50%). Thus, 33% x 50% = ~17%.

[NuclearSilo]

No, you got it wrong. I agree that the answer is 1/6. I just don't agree with other things.

other things being what?

Both case has the same probability of getting a number, but in reality, one method has more chance on getting that number than other method.

[Plutonium]

No, you got it wrong. I agree that the answer is 1/6. I just don't agree with other things.

other things being what?

Both case has the same probability of getting a number, but in reality, one method has more chance on getting that number than other method.

i doubt it. i learned that math is precise. if it says something, then it is.

but why dont u write a computer program to run a simulation of 1,000,000 trials to see if your theory holds? the simulation shouldn't take more than 1 second to complete.

or you can buy 6 marbles, and test your theory and do 1,000,000 trials and tell us the result u got in 42 years from today.